The main and auxiliary winding impedance of a 50-Hz, capacitor-start single-phase induction motor are Zm = (3 + j3) ohm and Za = (7 + j3) ohm. Determine the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of 90° between the currents of the two windings at start.
The main and auxiliary winding impedance of a 50-Hz, capacitor-start single-phase induction motor are Zm = (3 + j3) ohm and Za = (7 + j3) ohm. Determine the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of 90° between the currents of the two windings at start.
Right Answer is:
318 μF
SOLUTION
Given that ∅a + ∅m = 90°
$\begin{array}{*{20}{l}} {{{\tan }^{ – 1}}\left( {\dfrac{{{X_a} – {X_c}}}{{{R_a}}}} \right) – {{\tan }^ – }\left( {\dfrac{{{X_m}}}{{{R_m}}}} \right) = {{90}^ \circ }}\\ {{{\tan }^{ – 1}}\left( {\dfrac{{3 – {X_c}}}{7}} \right) – {{\tan }^ – }\left( {\dfrac{3}{3}} \right) = {{90}^ \circ }}\\ { \Rightarrow {{\tan }^{ – 1}}\left( {\dfrac{{3 – {X_c}}}{7}} \right) = 135}\\ { \Rightarrow \dfrac{{3 – {X_c}}}{7} = – 1} \end{array}$
⇒ Xc = 10
1/2πfC = 10
C = 318 µF